I think that I am going to have to have a very awkward conversation with my advisor next week.

He has proposed a rigorous definition of fractal, namely "A fractal is any set possessing non-real complex dimensions."

In $p$-adic land, singleton points possess non-real complex dimensions.

I don't think he is going to like that, as singleton points are not, and should not every be, fractals. :P

you're just looking for trouble

@XanderHenderson actually what is the dimension?

The complex dimensions of a point in $\mathbb{Q}_p$ are elements of the set $\mathbb{Z}\frac{2\pi i}{\log(p)}$

where $\mathbb{Z}a = \{ ka : k\in\mathbb{Z} \}$, as a set

with the complex dimensions living in $\mathbb{C}$ (hence "complex" dimensions)

and why?

why what?

why is it that dimension, I mean

P(X > 510) = 0.07

find z

@LeakyNun Do you mean using the Phi function?

ich weiss nicht, was du meinst die Phi Funktion

I have an other idea...It holds that Z = (X-mu)/sigma = (X-500)/sigma, right? But how can we apply here the given probability? @LeakyNun

finde sigma

du weisst das X 510 ist

Ok! Z = 10/sigma. But what is Z? @LeakyNun

die Solution von P(z>Z) = 0.07

"solution" heißt "Lösung"

@MatheinBoulomenos ich habe Vektorräume V und W über dem gleichen Körper, und eine lineare Abbildung von V nach W, das ich T heiße

*die Abbildung

warum die?

also ich meine, du brauchst "die" statt "das" als Relativpronomen

@LeakyNun I got stuck. Wie findet man diese Lösung?

@MaryStar by looking up a table

@MatheinBoulomenos oh right

@LeakyNun Ah the Standard normal table, right?

@LeakyNun okay, continue

ich will vergleichen diese Vektorräume: im T und im T*, wo T* : W* -> V* ist der Dual der lineare Abbildung, gegeben von w* |-> w* o T

@MaryStar yes

@LeakyNun The complex dimensions of a set are defined to be the poles of a certain meromorphic function associated to that set

Great! Thanks! @LeakyNun

@TedShifrin ist hier :D

Hi @Ted

@LeakyNun your word order is off, among other things

Hey @Ted

Nein, das ist er nicht.

hi demonic @Allesandro

hi, @Mathei: Glad to see you're a turning into a Deutschsprachlehrer :P

this function is defined by an integral which converges absolutely on the right half-plane bounded by $\{ \Re(s) = D\}$, where $D$ is the upper Minkowski dimension of the set $A$

ich will diese Vektorräume vergleichen: im T und im T*, wo T* : W* -> V* der Dual der lineare Abbildung ist, gegeben von w* |-> w* o T

under fairly mild hypotheses, this function will have a singularity at $D$, and so it recovers the Minkowski dimension

gegeben durch und der lineare_n_ Abbildung, aber die Reihenfolge ist jetzt korrekt ;)

which gives at least one justification for calling the set of poles "the complex dimensions"

@Xander: Is there a typo in that? Don't we accept certain real transcendental values as being Hausdorff dimensions of fractals?

Or maybe this is a different "dimension."

hi chat

ich will beweisen, das im T eine Teilmenge von im T* ist

metric spaces wierd things huh

@TedShifrin Where are you getting Hausdorff dimension from?

@LeakyNun das ergibt keinen Sinn

what other spaces should one know about ?

I myself said it. You're talking about some dimension I don't know about, I take it.

am taking first course in real analysis

ah

howdy Jacksoja

I am talking about the complex dimensions of a set

a set can have more than one complex dimension

I've never heard of such a thing unless I'm talking about algebraic varieties.

Hello Ted! @TedShifrin thanks again for that site!

@MatheinBoulomenos ich meine, das es ein Monomorphismus von im T nach im T* gibt

one of which will usually be the Minkowski dimension

@Jacksoja: If you master $\Bbb R$ and basic metric spaces you're doing just fine.

which is an upper bound on the Hausdorff dimension

Hey @Ted.

Hi @Balarka

I guess I don't know Minkowski dimension(s).

@TedShifrin What is so special about this metric space ? we only have a notion of distance right?

nothing rich appears in my eyes so far

Right ... But for analysis you want to estimate distances (for example, to talk about convergence, you want distance to go to 0).

basically, the (upper) Minkowski dimension asks how fast the volume of a ball scales relative to its radius

Just think about distance in $\Bbb R^n$ for starters. How do you tell if two vectors are close to one another?

relative complement is a contravariant functor

@LeakyNun Get out

lmao

Oh, but isn't that the foundational definition of Hausdorff dimension, ultimately, Xander?

well don't we use the triangle inequalit ?

|a-b|

as vectors

I'm gonna set this to my chat motto

Not triangle inequality.

The door is in that direction ----->

@Jacksoja You'll see why they are considered rich after learning about more general topological space which don't have nearly as many nice properties

@TedShifrin in what sense? The Hausdorff dimension gives a nice measure

Yes, @Jacksoja. So that's a fundamental example of metric space.

but it doesn't see oscillation in the geometry of a set

chat.stackexchange.com/users/201065#

done

I'm just saying that Hausdorff $d$-dimensional measure is built off that same notion, isn't it?

@AlessandroCodenotti all right ill take your words on that

kids these days

the complex dimensions, on the other hand, can be used to recover (for example) the spectrum of an "infinite harp"

How does one get a complex number from what you told me?

im actually quite proud of my memey chat description

its a multi-level meme

The Hausdorff and Minkowski dimensions have similar definitions, yes

@Balarka: Wiki entry coming for multi-meme?

lol

@MatheinBoulomenos for a second I thought you said "the high-school level" until I saw "exact sequence"

@Xander: So how can I get a complex number to be such a dimension? I'm intrigued.

@TedShifrin If $A$ is a bounded subset of $\mathbb{R}^N$, then we can associate to $A$ the function $$ \zeta_{A}(s) := \int_{A_{\delta}} d(x,A)^{s-N}\,\mathrm{d}x $$

$n=N$, I presume.

@MatheinBoulomenos the last equation makes no sense

where $A_\delta = \{ x : d(x,A) \le \delta \}$ is a $\delta$-neighborhood of $A$

Ah, and you're gonna do an analytic continuation of that zeta function.

yes indeed

Very interesting

but I never thought of using coker

and the poles of the zeta function are the complex dimensions

Got it. Why should I think of those as dimensions?

so, first off, you don't generally get just one dimension; you get several

Poles are indicator of how bad the set $A$ is I suppose

as to why, I don't really have a good argument, other than the fact that under very mild hypotheses, you recover the upper Minkowski dimension

That's a non-real dimension?

@LeakyNun the high level approach to your question: We have an exact sequence $W \to V \to K \to 0$, where $K= \operatorname{coker}(f)$ applying the dual space functor (which is contravariant and exact), we get an exact sequence $0 \to K^* \to V^* \to W^*$ now apply rank-nullity to get that $\operatorname{dim}(\operatorname{im}(f^*))= \operatorname{dim}(V^*) - \operatorname{dim}(K)= \operatorname{dim}(V)-\operatorname{dim}(W)+\operatorname{dim}(\operatorname{im}‌​(f))$

Ted I have a question about compact sets

indeed, the zeta function is absolutely convergent to the right of the Minkowski dimension, so the Minkowski dimension is typically teh abscissa of holomorphic convergence

@Xander: So far anything reasonable like a manifold, we don't pick up anything complex?

@MatheinBoulomenos why don't you need to dualize K in the new exact sequence?

my teacher defined it but was very vaig, kinda used the notion of a cover

@Jacksoja: Lots of people can answer such questions. Just ask :)

@TedShifrin For anything non-pathological, you will only get a single pole at the Minkowski dimension

That's the usual definition.

for more pathological sets, you can get periodic sets of poles

@LeakyNun fixed

Oh, I see. So you get a positive real pole.

I guess I should figure that out later.

So K subset of X is a compact set iff for every open covering of K there exist a finite subcover, that is the definition

right

$\zeta_A(s)$ is like the average distance-raised-to-some-power of $A$ from a $\delta$-neighborhood of it.

for example, the complex dimensions of the usual Ternary Cantor set are $\frac{\log(2)}{\log(3)} + \frac{2\pi i}{\log(3)} \mathbb{Z} $

is that well written ?

note that $\log_3(2)$ is the Hausdorff dimension of the Cantor set

So the poles should indicate upto what order things blow up as you get closer and closer to $A$

@MatheinBoulomenos so it should be $\dim(\operatorname{im}(f^*)) = \dim(V^*) - \dim(K^*)$ instead?

@Jacksoja: Where $X$ is your big metric space. Sure.

yeah, that's right, but every finite-dimensional vector space has the same dimension as it's dual

If $A$ is more space-filling, I expect the absolute value of some poles to be higher

Quite interesting, @Xander. I can't believe I've never been told about this before.

Like, take $A$ to be the Osgood curve or something

@MatheinBoulomenos when did I ever say the word "finite"?

Who is Osgood?

@TedShifrin Lapidus has been working on it for about a decade now ;)

evening everybody

Ah, long after he departed UGA.

@TedShifrin I shall return once I did what I need to do :) thanks as allways

@LeakyNun oh well

@TedShifrin That's the positive area Jordan curve on the plane

springer.com/us/book/9781461421757

and also springer.com/us/book/9783319447049

Gotcha, @Xander. I'm never buying another math book, but I appreciate the references.

heh

Springer was giving them away at JMM last week ;)

a fun problem

or, at least, one copy of FZF

hi @ted

Good stuff @Xander

salut, @Gabriel

Is the permanent of an upper triangular matrix equal to the multiplication of the numbers on the diagonal?

I wish I get around to learning more fractal geometry at some point

DO EET!

I quit AMS about 12 years ago and MAA when I retired ... hence persona non grata at JMM. :P

FRACTALS ARE LIFE!

heh

I'm sure someone could sneak you in if you really cared

I have Falconer flying around in my pdf reader

I met old friends for dinners, although there were plenty of other people it would have been fun to see.

@BalarkaSen Which Falconer? There are so many...

Uh, K. J.

no, I mean, which book

Falconer has written a lot of them

Geometry of fractal sets

ah

I did not know there were multiple books by him

that's a nice introduction

@MatheinBoulomenos so...

Yeah, there's Geometry of Fractal Sets, and Techniques something something, and a pretty blue Cambridge University Press volume named something, and a Short Introduction to Fractals from Oxford

jeez

but Geometry is a nice undergraduate-appropriate introduction

So I heard

@LeakyNun here's something you might like about dual spaces, for any subspace $U \subset V$, define $\operatorname{ann}(U) = \{f \in V^* \mid \forall u \in U f(u) = 0\}$ and for any subspace $W \subset V^*$ define $\operatorname{null}(W)= \{v \in V \mid \forall f \in W f(v)=0\}$. Then one can show that for any linear map $f:V \to W$ one has $\operatorname{ann}(\operatorname{im}(f))=\operatorname{ker}(f*)$ and $\operatorname{ann}(\operatorname{ker}(f)=\operatorname{im}(f^*)$

this feels very Galoisian to me

not at all ...

It's just the duality transcription of the usual stuff with transposes with matrices.

and $\operatorname{null}(\operatorname{ker}(f^*))=\operatorname{im}(f)$ and $\operatorname{null}(\operatorname{im}(f^*))=\operatorname{ker}(f)$

@Ted I don't need a single matrix to prove that :P

The row space is the image of $A^\top$ and its the orthogonal complement of the kernel of $A$, etc.

Nor do I, asshole.

@TedShifrin a lot of dualities can be written as Galois connections, so

its just a pain-in-the-ass description

orthogonal complement? what if my vector space doesn't have an inner product?

right... my battery is almost dead

later all

@Mathei: I'm not going to waste my time talking to you.

@MatheinBoulomenos but is im T a subspace of im T*?

roasted

more like, triggered

Galois connections per se are not really interesting... I don't consider a theory to be worthy of being a Galois theory without a Galois short exact sequence flying around somewhere

It's just a lot of categorical baggage

what's the Galois short exact sequence in the actual Galois theory?

@TedShifrin why are you so angry? I think there's a qualitative difference. If you have an inner product, then you have already chosen a way to identify $V$ and $V^*$, where the stuff I wrote above works without that

@LeakyNun If K/L/M is a sequence of Galois extensions 1 --> Gal(K/L) --> Gal(K/M) --> Gal(L/M) --> 1 holds

thx

that's just the third isomorphism theorem :P

The classic: K/M/K/L=L/M fractions cancel ses :P

well, short exact sequences are basically also the first isom. theorem hehe

they are first, but yeah

lol I literally meant to write first

my brain is only slowly coming out of being sick (or that's what my official policy will be for the next month if I make an error)

I agree that Galois connections are nothing deep. But yeah, if you look at subspaces (or even subsets) of $V$ and $V^*$, then the operations $\operatorname{ann}$ and $\operatorname{null}$ define an antitone Galois connection

This short exact sequence is the single most important fact that appears everywhere. In covering space theory you have $1 \to G \to \pi_1(X) \to \pi_1(X/G) \to 1$ (yes, that generalizes; I worked it out at some point but now I have forgotten). In group theory you have a left exact sequence at play that I don't really understand

I already told @Leaky about that

In Grothendieck's theory you also have the absolute Galois short exact sequence

1 --> Gal(Q) --> pi_1^et(X) --> pi_1^et(X^Q) --> 1 or something

yeah, I'd like understand that at some point

me too, but i'm a long way from there

why don't we just say that Gal is a contra-co-variant functor

because not all of us wants to fellate category theory

the answer to that question is no

why not?

how should it be? im T is a subspace of W and im T* is a subspace of V*

come on

I don't see a relation between the two

I mean, is there a monomorphism from im T to im T*

I see that as essentially identifying as a subspace

just like how S^1 is a subspace of S^2

It's not a canonical identification

true

which case are you talking about?

Hey @EricSilva

Forgot there were two erics now

Yo

Who's the other eric

I was thinking the latter, but also in the former case it's one where you can always identify subspace but at that point every space is a subspace of any larger dimensional subspace, so yeah

Eric Auld, who visited this chat earlier today

Ah cool

Did you see my thing about why $d\omega = 0$ should be an integrability condition from yesterday?

I feel like there should be a moving frame approach to proving Darboux now

Yeah

Oh ho ho

Now you're speaking my language

lmao. i don't know anything about your language but i feel it should be the right language

@LeakyNun lol, I applied did that thing wrong, sorry. We have $V \to W \to K \to 0$, where $K= \operatorname{coker}(f)$ applying the dual space functor (which is contravariant and exact), we get an exact sequence $0 \to K^* \to W^* \to V^*$ now apply rank-nullity to get that $\operatorname{dim}(\operatorname{im}(f^*))= \operatorname{dim}(W^*) - \operatorname{dim}(K)= \operatorname{dim}(W)-\operatorname{dim}(W)+\operatorname{dim}(\operatorname{im}‌​‌​(f))=\operatorname{dim}(\operatorname{im}‌​(f))$

this is basically an abstract nonsense proof that column rank = row rank

dim K*.

I said, it isn't finite

@EricSilva Ok, remind me, if I have a frame $(e_1, \cdots, e_n)$ along (a chart in) $M$, then the moving frames are given by taking the dual forms $(e^1, \cdots, e^n)$ on $T^* M$, letting $\omega = e^1 \otimes \cdots \otimes e^n$, and then the apparatus of forms $\Omega$ is the matrix-valued $1$-form such that $d\omega = \Omega \wedge \omega$?

Is that the abstract theory of moving frames?

A frame here of course means just a (local) section of the frame bundle $F(TM)/M$

@Leaky row rank = column rank is a matrix thing so there's not much of a point in talking about infinite dimensions

@MatheinBoulomenos Is that equivalent to this thing, if you replace orthocomplement by some vector space complement?

lol, covariant and contravariant

a tensor on a manifold is a section of $T M^{\otimes p} \otimes T^* M^{\otimes q}$. That's what it means.

If $q = 0$, covariant. If $p = 0$, contravariant. If both are nonzero, mixed

@BalarkaSen it's equivalent after you make some choices, yeah. Maybe I'm just being an algebraist who cares about unnecessary things, but I like my version because I don't need to make an arbitrary choice, like identifying a vector space with its dual

did I hear Balarka utter the word "contravariant"

@MatheinBoulomenos yeah right

dim K = dim K*?

I have uttered the word a lot of times. The word becomes dull after you actually learn to use it other than fellate it

Mornin

@LeakyNun I feel like you probably have a monomorphism in the infinite-dimensional case if you work with some cardinal arithmetic and proper analogs of rank-nullity in infinite dimensions. Since I don't know pretty much any non-trivial set theory, I won't try to do that

even if you assume the vector space to be finite

@MatheinBoulomenos Fair point

to say that dim K = dim K* is still to identify duals

it isn't

dimension is a numerical invariant, I'm not identifying anything

then why dim K = dim K*?

because Hom_F(F,F)=F and Hom_F(-,F) commutes with colimits

K?

oh yeah, I meant the ground field

@MatheinBoulomenos so?

Hi guys. Dumb question: How can I solve this inequality: $\frac{-x}{2}+\sqrt{\frac{x^2}{4}-y} \lt 0$ ?

Hom_F(F^n,F)=Hom(F,F)^n=F^n. I guess I need to use that every vector space is isomorphic to F^n, but you also need to do that to define dimension

$\sqrt{\frac{x^2}{4}-y} \lt \frac x 2$

$\frac{x^2}4 - y < \frac{x^2}4$

$y > 0$

you gotta be slightly careful

with what

it also has an additional restriction

what restriction?

@philmcole well of course you need to ensure that the square root makes sense

$x^2/4 -y >0 $ as well

or equal to 0

right

Ok thanks!

im going to do do some topology enjoy your wierd HOM's

cya

@Faust I've seen a lot of Homs in topology

@MatheinBoulomenos thats why im going to learn Topology cause i haven't :p

@BalarkaSen yes this is the idea (sorry I ran out)

Alright cool

Wow! Ted is pissed off.

Why is the Lusin-Theorem useful if the statement applies only for closed set?

I mean I could have an empty interior and the function is then nowhere continuous.

So there is no gained value on the continuity of the function. So every measurable function can still be very irregular.

@EricSilva How does one incorporate the Riemannian metric in this story for an intrinsically Riemannian manifold?

if my thing is embedded as a surface in R^3 I know that the $e^3$-component of $d\omega$ spits out the metric

Where $\{e_1, e_2, e_3\}$ is a frame with $e_3$ normal to the surface and 1, 2 are tangent frame along the surface

Also m8 this looks like the exterior covariant derivative operator to me. That's like $d_A \omega = d\omega + A \wedge \omega$ where $A$ is an End(E)-valued 1-form, isn't it?